Sketch the graph y=|x+3| evaluate integral …
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Sia ? 6 years ago
y = |x + 3|
{tex} \Rightarrow y = \left( {x + 3} \right){/tex}, {tex}if\ x\geq-3{/tex}
y = -(x + 3), if x < -3
{tex}\int_{ - 6}^0 {\left| {x + 3} \right|dx = ?} {/tex}
Area {tex}= \int_{ - 6}^{ - 3} {-\left( {x + 3} \right)dx + \int_{ - 3}^0 {\left( {x + 3} \right)dx} } {/tex}
{tex}= \left[ {-\frac{x^2}{2}-3x} \right]_{-6}^{-3}+ \left[ {\frac{x^2}{2}+3x} \right]_{-3}^{0}{/tex}
{tex}= \left[ {(-\frac{9}{2}+9)-(-\frac{36}{2}+18)} \right]+ \left[ {(0+0)-(\frac{9}{2}-9)} \right]{/tex}
{tex}= \left[ {(\frac{9}{2}+0)+(0+\frac{9}{2})} \right]{/tex}
= 9 sq units
2Thank You